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$ABC$ is an isosceles triangle with $AB=AC$, $\angle BAC=96^\circ$. $D$ is a point such that $\angle DCA=48^\circ$, $AD=BC$ and angle $DAC$ is obtuse. What is the measure (in degrees) of $\angle DAC$?

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  • $\begingroup$ Please do not delete questions with good answers. Berci devoted effort to answer your question and deserves the chance to get rewarded for it. Others may also benefit from your question and its answers. $\endgroup$
    – robjohn
    Commented Feb 21, 2013 at 13:49
  • $\begingroup$ This is a problem posted on Brilliant.org, which provides weekly problem sets. You may view it here. I request that this is closed off for discussion till Monday. - Calvin Lin, Brilliant Challenge Master. $\endgroup$
    – Calvin Lin
    Commented Feb 21, 2013 at 15:17

2 Answers 2

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First, draw. Then calculate the angles you might already know and name the unknown ones. So, using that $ABC$ is isosceles, we have $\angle BAC=42^\circ$. Since $42+48=90$, we have $\cos 48^\circ=\sin 42^\circ$. Let $\delta:=\angle ADC$, which is $<90^\circ$ since $\angle DAC$ is obtuse. After finding $\delta$, we will have $$\angle DAC=180^\circ-48^\circ-\delta\ .$$

Use the theorem of sines, for the triangles $ABC$ and $ACD$, and use the identity for $\sin(2x)$ with $x=48^\circ$:

$$\frac{\sin\delta}{\sin 48^\circ}=\frac{AC}{AD}=\frac{AC}{BC}= \frac{\sin 42^\circ}{\sin 96^\circ}$$

From this, as $\sin 96^\circ=2\sin 48^\circ\cos 48^\circ=2\sin 48^\circ\sin 42^\circ $, we get that $\sin\delta=1/2$, that is $\delta=30^\circ$.

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  • $\begingroup$ I have another solution. Can u point out my mistake if my solution is incorrect. Here is the full working: img109.imageshack.us/img109/9035/triangleg.png $\endgroup$
    – user62947
    Commented Feb 19, 2013 at 12:55
  • $\begingroup$ From the picture I can read that you are aware of everything for this problem. A minor mistake is that your result is (not much but) less than $90^\circ$, though should be bigger. The main mistake is that the original text says $AD=BC$ whereas in your picture $CD=BC$. All else is fine and correct. $\endgroup$
    – Berci
    Commented Feb 19, 2013 at 13:52
  • $\begingroup$ This is a problem posted on Brilliant.org, which provides weekly problem sets. You may view it here. I request that this is closed off for discussion till Monday. - Calvin Lin, Brilliant Challenge Master. $\endgroup$
    – Calvin Lin
    Commented Feb 21, 2013 at 15:18
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Take $O$ the circumcenter of $\triangle ADC$, see that $\widehat{DAO}=\widehat{ADO}=\widehat{ABC}=\widehat{ACB}=42^\circ$, so $\triangle ADO\cong\triangle BCA$, a.s.a., thus $AO=CO=AC$ and $\triangle AOC$ is equilateral, i.e. $\widehat{ADC}=30^\circ, \widehat{DAC}=102^\circ$.

Best regards!

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